This is a problem that has been given to me for homework:
$ABCD$ is a tetrahedron such that $AD=BD$ and $AC=BC$. The angle between $CD$ and plane $ABC$ is $75^\circ$ and the perimeter of $\triangle ABC$ is $3$. Find the maximum volume of the tetrahedron if its' circumcenter lies on plane $ABC$.
Here's what I've tried:
So I first let $CH$ be the altitude from $C$ in $\triangle ABC$ and $DO$ be the altitude from $D$ in the tetrahedron. Obviously, $O\in CH$. Then I denote $K$ the intersection point of $CH$ and the sphere. Since it's center lies on plane $ABC$, $\angle KDC =90^\circ$. We have $\sphericalangle DCO=75^\circ$ and therefore $DO=\dfrac{1}{4}KC$. However I don't know how to use the given perimeter of $\triangle ABC$. Any ideas?
Denote $DO=h$ so that the diameter of the circumscribed circle about $\triangle ABC$ is $KC=4h$. Circumradius of $\triangle ABC$ is thus equal to $8h$ as it is twice less than $KC$. I'm going to use the standard notations for the sides of $\triangle ABC$ $-$ $a$, $b$ and $c$. Its' area is equal to $\dfrac{abc}{8h}$ and now for the volume $V$ of the tetrahedron we obtain $\dfrac{abc}{24}$. This is where the perimeter of the triangle comes in handy. By the AM-GM inequality, we have $\dfrac{a+b+c}{3}=1\geq\sqrt[3]{abc}$ and so $abc\le 1$. From here we have $V\le\dfrac{1}{24}$ so the answer to your problem is $\dfrac{1}{24}$.